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DIFFERENTIAL

EQUATIONS

Basic Terminology

A differential equation is an equation

that contains an unknown function to-

gether with one or more of its deriva-

tives.

1

Examples:

1. y′ = 2x + cos x

2.dy

dt= ky (exponential growth/decay)

3. x2y′′ − 2xy′ + 2y = 4x3

2

4.∂2u

∂x2+

∂2u

∂y2= 0 (Laplace’s eqn.)

5.d3y

dx3− 4

d2y

dx2+ 4

dy

dx= 0

3

TYPE:

If the unknown function depends on a

single independent variable, then the

equation is an

ordinary differential equation (ODE);

if the unknown function depends on

more than one independent variable,

then the equation is a

partial differential equation (PDE).

4

ORDER:

The order of a differential equation is

the order of the highest derivative of

the unknown function appearing in the

equation.

5

Examples:

1. y′ = 2x + cos x

2.dy

dt= ky (exponential growth/decay)

3. x2y′′ − 2xy′ + 2y = 4x3

6

4.∂2u

∂x2+

∂2u

∂y2= 0 (Laplace’s eqn.)

5.d3y

dx3− 4

d2y

dx2+ 4

dy

dx= 0

6.d2y

dx2+2x sin

(

dy

dx

)

+3exy =d3

dx3(e2x)

7

SOLUTION:

A solution of a differential equation

is a function defined on some domain

D such that the equation reduces to

an identity when the function is substi-

tuted into the equation.

8

Examples:

1. y′ = 2x + cos x

9

2. y′ = ky

Set of all solutions

y = Cekt, C any constant

10

3. x2y′′ − 2xy′ + 2y = 4x3

Is y = x2 + 2x3 a solution?

11

4. x2y′′ − 2xy′ + 2y = 4x3

Is y = 2x + x2 a solution?

12

Note:

y = C1x + C2x2 + 2x3

is a solution for any numbers C1, C2.

And, this is the set of all solutions of

x2y′′ − 2xy′ + 2y = 4x3

13

5.∂2u

∂x2+

∂2u

∂y2= 0

u = ln√

x2 + y2 Solution?

14

6.∂2u

∂x2+

∂2u

∂y2= 0

u = cos x sinh y, u = 3x − 4y

Solutions??

15

6. Find a value of r, if possible,

such that y = erx is a solution of

y′′ − 2y′ − 15y = 0.

16

7. Find a value of r, if possible,

such that y = xr is a solution of

x2y′′ − 4x y′ + 6y = 0.

17

8. Find a value of r, if possible,

such that y = xr is a solution of

y′′ −1

xy′ −

8

x2y = 0.

18

From now on, all differential equa-

tions are ordinary differential equa-

tions.

19

n-PARAMETER FAMILY OF SO-

LUTIONS / GENERAL SOLUTION

Example: Solve the differential equa-

tion:

y′′′ − 12x + 6e2x = 0

20

NOTE: To solve a differential equa-

tion having the special form

y(n)(x) = f(x),

simply integrate f n times,

AND each integration step produces an

arbitrary constant..

21

Intuitively, to find a set of solutions of

an n-th order differential equation

F[

x, y, y′, y′′, . . . , y(n)]

= 0

we “integrate” n times, with each in-

tegration step producing an arbitrary

constant of integration (i.e., a param-

eter). Thus, ”in theory,” an n-th order

differential equation has an n-parameter

family of solutions.

22

SOLVING A DIFFERENTIAL EQUA-

TION:

To solve an n-th order differential equa-

tion

F (x, y, y′, y′′, . . . , y(n) = 0

means to find an n-parameter family of

solutions. (Note: Same n.)

NOTE: An “n-parameter family of

solutions” is more commonly called the

GENERAL SOLUTION.23

Examples: Find the general solution:

1. y′ − 3x2− 2x + 4 = 0

24

2. y′′ − 2x + sin 2x = 0

25

3. y′′′ − 3y′′ + 3y′ − y = 0

Answer: y = C1ex + C2xex + C3x2ex

4. x2y′′ − 2xy′ + 2y = 4x3

Answer: y = C1x + C2x2 + 2x3

26

PARTICULAR SOLUTION:

If specific values are assigned to the

arbitrary constants in the general solu-

tion of a differential equation, then the

resulting solution is called a particular

solution of the equation.

27

Examples:

1. y′′ = 6x + 8e2x

General solution:

y = x3 + 2e2x + C1x + C2

Particular solutions:

28

2. x2y′′ − 2xy′ + 2y = 4x3

General solution:

y = C1x + C2x2 + 2x3

Particular solutions:

29

SINGULAR SOLUTIONS:

Solutions of an n-th order differential

equation which are not included in an

n-parameter family of solutions (i.e., in

the general solution) are called singu-

lar solutions.

30

Example: y′ = 4x√

y − 2

General solution:

√

y − 2 = (x + C)2

Note: y = 2 is also a solution AND it

is NOT a particular solution.

Set of all solutions:

√

y − 2 = (x + C)2, & y ≡ 2

31

THE DIFFERENTIAL EQUATION

OF AN n-PARAMETER FAMILY:

Given an n-parameter family of curves.

The differential equation of the fam-

ily is an n-th order differential equation

that has the given family as its general

solution.

32

Examples:

1. y2 = Cx3 + 4 is the general solu-

tion of a DE.

a. What is the order of the DE?

b. Find the DE?

33

2. y = C1x + C2x3 is the general

solution of a DE.

a. What is the order of the DE?

b. Find the DE?

34

General strategy for finding the dif-

ferential equation

Step 1. Differentiate the family n

times. This produces a system of n+1

equations.

Step 2. Choose any n of the equa-

tions and solve for the parameters.

Step 3. Substitute the “values” for

the parameters in the remaining equa-

tion.35

Examples:

The given family of functions is the

general solution of a differential equa-

tion.

(a) What is the order of the equation?

(b) Find the equation.

36

1. y = Cx3− 2x

(a)

(b)

37

2. y = C1e2x + C2e3x

(a)

(b)

38

3. y = C1 cos 3x + C2 sin 3x

(a)

(b)

39

4. y = C1x4 + C2x + C3

(a)

(b)

40

5. y = C1 + C2x + C3x2

(a)

(b)

41

INITIAL-VALUE PROBLEMS:

1. Find a solution of

y′ = 3x2 + 2x + 1

which passes through the point (−2,4).

42

2. y = C1 cos 3x + C2 sin 3x is the

general solution of

y′′ + 9y = 0.

a. Find a solution which satisfies

y(0) = 3

43

b. Find a solution which satisfies

y(0) = 4, y(π) = 4

44

c. Find a solution which satisfies

y(0) = 3, y′(0) = 4

45

An n-th order initial-value problem

consists of an n-th order differential

equation

F[

x, y, y′, y′′, . . . , y(n)]

= 0

together with n (initial) conditions of

the form

y(c) = k0, y′(c) = k1, y′′(c) = k2, . . . ,

y(n−1)(c) = kn−1

where c and k0, k1, . . . , kn−1 are

given numbers.

46

NOTES:

1. An n-th order differential equation

can always be written in the form

F[

x, y, y′, y′′, · · · , y(n)]

= 0

by bringing all the terms to the left-

hand side of the equation.

2. The initial conditions determine

a particular solution of the differential

equation.

47

Strategy for Solving an Initial-Value

Problem:

Step 1. Find the general solution of

the differential equation.

Step 2. Use the initial conditions to

solve for the arbitrary constants in the

general solution.

48

Examples:

1. Find a solution of the initial-value

problem

y′ = 4x + 6e2x, y(0) = 5

49

2. y = C1e−2x + C2e4x is the general

solution of

y′′ − 2y′ − 8y = 0

Find a solution that satisfies the initial

conditions

y(0) = 3, y′(0) = 2

50

3. y = C1x + C2x3 is the general

solution of

y′′ −3

xy′ +

3

x2y = 0

a. Find a solution which satisfies

y(1) = 2, y′(1) = −4.

51

b. Find a solution which satisfies

y(0) = 0, y′(0) = 2.

c. Find a solution which satisfies

y(0) = 4, y′(0) = 3.

52

EXISTENCE AND UNIQUENESS:

The fundamental questions in a course

on differential equations are:

1. Does a given initial-value problem

have a solution? That is, do solutions

to the problem exist ?

2. If a solution does exist, is it

unique ? That is, is there exactly one

solution to the problem or is there more

than one solution?53